Problem: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{-10n^3 - 40n^2 + 120n}{n^3 + 2n^2 - 24n}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {-10n(n^2 + 4n - 12)} {n(n^2 + 2n - 24)} $ $ k = -\dfrac{10n}{n} \cdot \dfrac{n^2 + 4n - 12}{n^2 + 2n - 24} $ Simplify: $ k = - 10 \cdot \dfrac{n^2 + 4n - 12}{n^2 + 2n - 24}$ Since we are dividing by $n$ , we must remember that $n \neq 0$ Next factor the numerator and denominator. $ k = - 10 \cdot \dfrac{(n + 6)(n - 2)}{(n + 6)(n - 4)}$ Assuming $n \neq -6$ , we can cancel the $n + 6$ $ k = - 10 \cdot \dfrac{n - 2}{n - 4}$ Therefore: $ k = \dfrac{ -10(n - 2)}{ n - 4 }$, $n \neq -6$, $n \neq 0$